Question 1206065
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The standard setup for solving the problem algebraically....:<br>
x = amount that yielded 7% interest
20000-x = amount that yielded 11% interest<br>
The total interest earned was $1560:<br>
{{{.07(x)+.11(20000-x)=1560}}}<br>
I'll leave it to you to finish solving the problem by that method, using basic algebra.<br>
A quick informal method for solving any 2-part "mixture" problem like this, if a formal algebraic solution is not required....<br>
(1) $1560 interest on an investment of $20,000 is a return of 1560/20000 = 0.078 = 7.8%.
(2) Using a number line if it helps, observe/calculate that 7.8% is 0.8/4.0 = 1/5 of the way from 7% to 11%.
(3) That means 1/5 of the total was invested at the higher rate.<br>
1/5 of $20,000 is $4000.<br>
ANSWER: $4000 was invested at 11%; the other $16,000 was invested at 7%.<br>
CHECK: .11(4000)+.07(16000) = 440+1120 = 1560<br>